anonymous
  • anonymous
lim [sqrt(x^2+12)-4]/(x-2) x>2 Can someone show me how it's 1/2?
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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zzr0ck3r
  • zzr0ck3r
\[\lim_{x\rightarrow2}\frac{\sqrt{x^2+12}-4}{x-2}?\]
anonymous
  • anonymous
yes
zzr0ck3r
  • zzr0ck3r
brb phone.

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More answers

Luigi0210
  • Luigi0210
Well.. while we wait for zz
Luigi0210
  • Luigi0210
Do you know L'hopital?
anonymous
  • anonymous
Yeah I do
anonymous
  • anonymous
That's makes things easy.
Luigi0210
  • Luigi0210
Yea, try it out
anonymous
  • anonymous
Try that. Might make it easier.
anonymous
  • anonymous
I'm not really experienced using it with square roots. I get sqrt2x/1
Luigi0210
  • Luigi0210
Do you know the derivative of a square root?
anonymous
  • anonymous
Nope
anonymous
  • anonymous
|dw:1378172628590:dw|
anonymous
  • anonymous
a' is the derivative of a.
Luigi0210
  • Luigi0210
What dido said :P
zzr0ck3r
  • zzr0ck3r
@Bleary0mite you getting it?
anonymous
  • anonymous
Not really, to be honest.
zzr0ck3r
  • zzr0ck3r
\[\frac{d}{dx}(\sqrt{x^2+12})=\frac{d}{dx}(x^2+12)^{\frac{1}{2}}=\frac{1}{2}(x^2+12)^{-\frac{1}{2}}*\frac{d}{dx}(x^2+12)\]
zzr0ck3r
  • zzr0ck3r
does this make sense?
anonymous
  • anonymous
Oh okay that makes more sense to me
zzr0ck3r
  • zzr0ck3r
bring the 1/2 down the de-increment 1/2 down by 1 and get -1/2
zzr0ck3r
  • zzr0ck3r
and multiply by the derivative of the inside
zzr0ck3r
  • zzr0ck3r
so now we have\[=\frac{2x}{2\sqrt{x^2+12}}\]
zzr0ck3r
  • zzr0ck3r
now this is just the top\[=\frac{x}{\sqrt{x^2+12}}\]and if we take the derivative of the bottom we have1
zzr0ck3r
  • zzr0ck3r
you with me?
anonymous
  • anonymous
Yeah I follow
zzr0ck3r
  • zzr0ck3r
\[\lim_{x\rightarrow2}\frac{\frac{x}{x}}{\frac{\sqrt{x^2+12}}{x}}=\lim_{x\rightarrow2}\frac{1}{\sqrt{\frac{x^2}{x^2}+\frac{12}{x^2}}}\]now plug in 2\[=\frac{1}{\sqrt{1+\frac{12}{4}}}=\frac{1}{\sqrt{4}}=\frac{1}{2}\]
zzr0ck3r
  • zzr0ck3r
make sense?
anonymous
  • anonymous
Why did we have to divide everything by x?
zzr0ck3r
  • zzr0ck3r
its a trick to use with limits its sort of like this \[\lim_{x\rightarrow \infty}\frac{2x+3}{3x+5}\]divide top and bottom by x \[\lim_{x\rightarrow \infty}\frac{\frac{2x}{x}+\frac{3}{x}}{\frac{3x}{x}+\frac{5}{x}}=\lim_{x\rightarrow \infty}\frac{2+\frac{3}{x}}{3+\frac{5}{x}}\]now we run the limit\[\frac{2+\frac{2}{\infty}}{3+\frac{5}{\infty}}=\frac{2+0}{3+0}=\frac{2}{3}\]
zzr0ck3r
  • zzr0ck3r
I guess the answer to your question is why not:)
zzr0ck3r
  • zzr0ck3r
notice that x is not 0, so we can divide by it. and it makes things easy.
anonymous
  • anonymous
Alright yeah I see that. I wasn't aware of that trick. thanks
zzr0ck3r
  • zzr0ck3r
o wow im silly lol we could have just plugged in x.=2 \[\lim_{x\rightarrow 2}\frac{x}{\sqrt{x^2+12}}\]plug in 2\[\frac{2}{\sqrt{4+12}}=\frac{1}{2}\]
zzr0ck3r
  • zzr0ck3r
but now you know about the trick:)
anonymous
  • anonymous
YES! that's exactly why I asked why we divided by x lol
zzr0ck3r
  • zzr0ck3r
im leaving to Paris in a few hours...sorry its been a nutty day
anonymous
  • anonymous
Alright I appreciate your help. Thank you
zzr0ck3r
  • zzr0ck3r
imagine*
zzr0ck3r
  • zzr0ck3r
its a very usefull trick and most people dont use it enough:)
zzr0ck3r
  • zzr0ck3r
np
anonymous
  • anonymous
I'll keep it in mind
anonymous
  • anonymous
Here is a step by step solution: http://www.symbolab.com/solutions/limits?query=%5Clim_%7Bx%5Cto2%20%7D%5Cleft(%5Cfrac%7B%5Csqrt%7Bx%5E%7B2%7D%2B12%7D-4%7D%7Bx-2%7D%5Cright)

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